A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter8: Data And Network Communication Technology

Section: Chapter Questions

Problem 6RQ

See similar textbooks

Related questions

Q: 22: In PSK system the bit rate of NRZ stream is 20 Mbps and carrier frequency is 2GHZ. Find the…

A: Answer is given below-

Q: Ten signals, each requiring 4000 Hz, are multiplexed onto a single channel using FDM. What is the…

A: The answer is

Q: ) Hexadecimal byte of Ox58 is passed to the physical layer of the network. Assuming, Manchester…

A: Below i am using manchester encoding :…

Q: A channel has a bit rate of 4 kbps and a propaga- tion delay of 20 msec. For what range of frame…

A: The Answer is

Q: eam of data is being carried by STS-1 frames. If the data rate of the stre 40 Mbps, how many STS-1…

A: Introduction: The data rate is the number of bits transmitted per second from one device to another…

Q: A path in a digital circuit-switched network has a data rate of 1 Mbps. The exchange of 1000 bits is…

A: Here, we have to write solutions for the above questions.

Q: Assume two hosts are paired in a CSMA/CD network with a 1Gbps of transmission medium. How far apart…

A: Given, Bandwidth (B) = 1 GBPS Length (L) = 1,000 bits Velocity (V) = 2*108m/s Here we have to…

Q: Hexadecimal byte of Ox58 is passed to the physical layer of the network. Assuming, Manchester…

A: Manchester encoding (first published in 1949) is a synchronous clock-encoding technique used by the…

Q: A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the…

A: The Answer is

Q: what is the throughput if the system produces the 1000 frames/sec then A slooted ALOHA network…

A: The slotted ALOHA calculation is

Q: Consider the following scenario: Frames of 12000 bits are sent over a 400kbps channel using a…

A: Protocols in Networking A network protocol is a set up set of decides that decide how information…

Q: Consider a 50 kbps satellite channel with a 500 milliseconds round trip propagation delay. If the…

A: Here in this question we have given a satallite with bandwidth 50kbps. RTT =500 ms. Frame length…

Q: Suppose you are designing a sliding window GBN protocol for a 5 Mbps point to point link, which has…

A: Find the required answer with calculation given as below :

Q: A 3000km long trunk operates at 1.536 Mbps is used to transmit 64 byte frames. If the propagation…

A: The Answer is

Q: 12. A channel has a data rate of R = 1 [Kbps] and a propagation delay of t= 0.2 [sec/km]. The…

Q: what is the throughput if the system produces the 1000 frames/sec then A pure ALOHA network…

A: throughput of ALOHA network is calculated in step 2.

Q: A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each…

A: Introduction serial transmission T1 info bits 8 Start bit 2 Stop bot 1 Parity bit 1 synchronous…

Q: A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on…

A: ALOHA: ALOHA originally stood for Additive Links On-line Hawaii Area. In the 1970s ALOHA random…

Q: what is the throughput if the system produces the 250 frames/sec then A slooted ALOHA network…

A: The Slotted ALOHA calculation is

Q: III. The figure below shows synchronous TDM, the data rate for each input connection is 1 Mbps. If 1…

Q: QI/ A DS-CDMA system with only two senders-receiver pairs with the following information, show the…

A: Answer: I have given answered in the handwritten format in brief explanation

Q: Frames of 1000 bits are sent over a 1 Mbps link between earth and moon whose propagation delay is…

A: The Answer is

Q: Suppose you are designing a sliding window protocol for a 2Mbps point to point link to the moon,…

A: According to the information given:- We ahve to find minimum number of bits you need for the…

Q: a) A pure ALOHA channel is going to be shared among some number of stations. Each station is…

A: Actually, 1 byte =8 bits. given each station is transmitting a 1024 -byte frame average every 28…

Q: In asynchronous transmission, why does the gap time between bytes differ?

A: Introduction: Because Transmission is asynchronous, the time interval between bytes is "variable."

Q: Consider a baseband bus where the average distance between any two stations is 375m with a data rate…

A: As given - Distance between two stations = 375m Data rate = 10Mbps Propagation delay = 200m/us…

Q: Assume having a low-pass noiseless channel that spans from a Frequency of 0Hz to 130Hz, to send at a…

A: please upvote for my efforts thanks! answer: bit rate= 2* bandwidth * log2(number of levels)

Q: A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and l parity bit for each…

A: Introduction :We have given two transmission T1, T2 where T1 is a serial transmissionAnd T2 is a…

Q: Find the new durations for a) the input bit duration, b) the output bit duration, c bit rate, and d)…

A: When there are four channels then each frame will carry 2 bytes from each Channel A)Therefore 2byte…

Q: A 4480-octet datagram is to be transmitted and needs to be fragmented because it will pass through…

A: given: 4480-octet datagram maximum payload of 1500 octets To find: Total Length, More Flag, and…

Q: If four 4-kbps connections are multiplexed together using synchronous TDM. Find the following 1. The…

A: 1) The duration of 1 bit for each connection is 1/4kbps =0.00025s = 0.25milliseconds2)transmission…

Q: II. The figure below shows synchronous TDM, the data rate for each input connection is 1 Mbps. If 1…

A: Solution:-

Q: what is the throughput if the system produces the 500 frames/sec then A pure ALOHA network transmits…

A: The Answer is

Q: A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the…

A: Actually, Network is consists of two or more computers that are linked in order to share resources…

Q: what is the throughput if the system produces the 250 frames/sec then A pure ALOHA network transmits…

A: The answer is in step 2:

Q: Consider that the link capacity of a channel is 512 Kbps and round – trip delay time is 1000ms find…

A: Given: channel is 512 Kbps delay time is 1000ms

Q: Five channels, each with a 100 kHz bandwidth, are to be multiplexed together What is the minimum…

A: Five channels, each with a 100 kHz bandwidth, are to be multiplexed together What is the minimum…

Q: The distance from earth to a distant planet is approximately 9 x 10^10 m. What is the channel…

A: Answer the above questions are as follows:

Q: An RF frequency has 30-36 MHz as its lower andhigher frequencies. It has a current channel capacity…

Q: Consider TDM-based circuit switching with 8000 frames per second and 32 time slots per frame. During…

A: Circuit Switching:- In circuit-switched networks, the resources needed along a path(buffers, link…

Q: Computer A, B C, D and E is arranged in a ring topology. Calculate the propagation delay of sending…

A: All workings are below Explanation: Detailed explanation:

Q: Consider sliding window protocol for 80Mbps point - to - point link with propagation delay of 2 sec.…

A: Given, Bandwidth = 80 Mbps Frame Size = 4 KB = 4096 * 8 bits = 32768 bits Propagation delay = tp = 2…

Q: Assume having a low-pass noiseless channel that spans from a Frequency of OHz to 180HZ, to send at a…

A: please do upvote for my efforts ! answer: bit rate = 2* bandwidth * log2(number of levels)

Q: n rule, determine the system bandwidth that can transmitted imum bit rate of 10 Gbps when the signal…

Q: Suppose a 128-Kbps point-to-point link is set up between Earth and a rover on Mars. The distance…

A: Answer is in next step.

Q: Consider a CSMA/CD with frame size is 20 KB and transmission rate is 10 Kbps, then maximum length of…

A: Introduction :

Q: Five channels, each with a 400-KHz bandwidth, are to be multiplexed together. What is the minimum…

A: In a communication channel, the transmission media have multiple channels in which a single channel…

Q: what is the throughput if the system produces the 500 frames/sec then A slooted ALOHA network…

A: The slotted Aloha calculation is

Question

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Trending now

This is a popular solution!

Step by step

Solved in 4 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Similar questions

A planet located far from Earth is around 9 x 1010 meters away. If a stop-and-wait protocol is employed for frame transmission on a 64 Mbps point-to-point link, what is the channel utilization? Assume that the speed of light is 3 108 m/s and that the frame size is 32 KB.
Consider a full-duplex 256 Kbps satellite link with a 240 millisecond end-to-end delay. The data frames carry a useful payload of size 12,000 bits. Assume that both ACK and data frames have 320 bits of header information, and that ACK frames carry no data. What is the effective data throughput When using Stop and Wait?
The distance from earth to a distant planet is approximately 9 x 10^10 m. What is the channel utilization if a stop-and-wait protocol is used for frame transmission on a 64 Mbps point-to-point link? Assume that the frame size is 32 KB and the speed of light is 3 × 10^8 m/s.
Que.a) Consider a CSMA/CD network that transmits data at a rate of 400 x 106 bits per second. The maximum signal propagation time from one node to another node is 20 microseconds. The minimum size of a frame in the network is(. bytes).
Topic :- Data link Layer A channel has a bit rate of 4 kbps and a propaga tion delay of 20 msec. For what range of frame sizes does stop-and-wait given an efficiency of at least 50 percent? Note:- Final Answer L= 5.33× 10-7 kbps Give step by step answer
Consider a CSMA/CD with frame size is 20 KB and transmission rate is 10 Kbps, then maximum length of the cable (km) required, if signal speed is 15 km/sec.
Consider a full-duplex 512 kbps link with length 36,000Km connecting two hosts. The data frames carry a useful payload of size 6000 bits. Assume that both ACK and data frames have 400 bits of header information, and that ACK frames carry no data. The propagation speed along this link is 3×108 m/sec. a) What is the RTT on this link? b) What is the effective useful data throughput, in kbps, when using Stop-and-Wait?
8. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps.What is the throughput if the system (all stations together) produces?a. 1000 frames per second?b. 500 frames per second?c. 250 frames per second?
Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25 ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link). 1.1 What is the minimum number of bits (1) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
Assume that a channel occupies a bandwidth of 110 kHz. We need to combine eleven channels into a link with a bandwidth of 1.26 MHz, from 1 to 2.26 MHz with 5 kHz guard band. Show the configuration, using the frequency domain.
Please check my answer and correct the part which is need to be foxed.And provide justification for the answer. a) The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.0025ms Transmission time (Tt) = frame size/bandwidth = 10000 * 8bits / (1Mbps * 10 ^ 6) = 0.008sec = 8ms Total time for transmission = 2 * Tp + Tt(since the frame has to travel from A to D and then D to A) = 2 * 0.0025ms + 8ms = 0.005ms + 8ms = 8.005ms The transmission time of the frame from A to D is 8.005 ms. b) The efficiency of the CSMA/CD protocol is given by the formula: Tt is the transmission time of a frame C is the number of collisions, Tp is the propagation time of a signal from one end of the segment to the other. Efficiency = Tt / (C * 2 * Tp + Tt + Tp) Assuming that there are no other stations transmitting or attempting to transmit on the segment, the transmission from A to D will succeed without collisions. Therefore,…
Ten signals, each requiring 4000 Hz, are multiplexed onto a single channelusing FDM. What is the minimum bandwidth required for the multiplexed channel? Assume that theguard bands are 400 Hz wide.